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Match List  I (Roots in splane) with List  II (Corresponding impulse response) and select the correct answer using the codes given below the lists:
List  I
List  II
Codes:
For A: Two conjugate poles on imaginary axis i.e. it’s impulse response = sin at (marginally stable).
For B: Two poles at origin i.e. impulse response = t u(t) (unstable system).
For C: Two complex conjugate poles with negative real parts. So, its impulse response = er^{at} sin bt (stable system).
For D: Two imaginary complex double roots. So, impulse response = t sin bt. (unstable system).
Transfer function is applicable to only linear and timevariant systems.
The DC gain of the system represented by the following transfer function is
Given transfer function is
Converting above transfer function in timeconstant form, we get:
Hence, dc gain is
The unit impulse response of a control system is given by c(t) = te^{t} + e^{t}. Its transfer function is
Given, c(t) = te^{t} + e^{t} (impulse response)
We know that,
The impulse response of an initially relaxed linear system is e^{3t} u(t). To produce a response of te^{3t} u(t), the input must be equal to
We know that, impulse response
For given system,
Now, c(t) = t e^{3t}
So, R{s) = C(s) x (s + 3)
or, r(t) = Required input = e^{3t} u(t)
A linear time invariant system, initially at rest when subjected to a unit step input gave response c(t) = te^{t}(t ≥ 0). The transfer function of the system is
Given, c(t) = te^{t}
Now, r(t) = u(t) (unit step input)
∴ R(s) = 1/s
So,
A linear time invariant system having input r(t) and output y(t) is represented by the differential equation
The transfer function of the given system is represented as
Taking Laplace transform of the given differential equation on both sides, we get
2s^{2} Y(s) + sY(s) + 5 Y(s) = R(s) + 2e^{s} R(s)
or,
The open loop transfer function of a system is
A closed loop pole will be located at s = 12
when the value of K is
The singularities of a function are the points in the splane at which the function or its derivates
Assertion (A): The final value theorem cannot be applied to a function given by
Reason (R): The function s F(s) has two poles on the imaginary axis of splane.
The final value theorem is valid only if sF(s) does not has any poles on the jω axis and in the right half of the splane. Hence, both A and R are true and R is a correct explanation of A.
When two time constant elements are cascaded noninteractively then, the overall transfer function of such an arrangement
The unit impulse response of a unity feedback control svstem whose Open Iood transfer function
Given, O.L.T.F.
For unit impulse response, R(s) = 1
or, c(t) = 2(e^{t } te^{t}) + te^{t}
= 2e^{t}  te^{t} = (2 t) e^{t}
Thus, impulse response, c(t) = (2  t) e^{t} u(t)
The impulse response of a linear time invariant system is a unit step function. The transfer function of this system would be
Impulse response,
c(t) = u(t)
Consider the function where F(s) is the Laplace transform of f(t) is equal to
Using final value theorem
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